Confused because $\frac{d}{dx}\arctan(x)=-\frac{d}{dx}\mathrm{arccot}(x)$

Question

I was surprised to discover that

$$\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$$

$$\frac{d}{dx}\arccos(x)=\frac{-1}{\sqrt{1-x^2}}$$

This would seem to imply $\arcsin=-\arccos$ up to a constant, but then I discovered the identity

$$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$

and then it all made sense.

As a consequenc, if I am integrating $\frac{\pm 1}{\sqrt{1-x^2}}$ I can arbitrariy pick either $\arcsin$ or $\arccos$ as an antiderivative. So far so good.

I then discovered that, similarly,

$$\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}=-\frac{d}{dx}\mathrm{arccot}(x)$$

and assumed the explanation would be the same. But I sanity-checked it, and it turns out

$$\arctan(x)+\mathrm{arccot}(x)=\pm\frac{\pi}{2}$$

is not a constant.

So, if I am integrating $\frac{1}{1+x^2}$, how do I pick an antiderivative?

Answer 1

The derivatives are $$\DeclareMathOperator{\arccot}{arccot} \arctan'x=\frac{1}{1+x^2} \qquad \arccot'x=-\frac{1}{1+x^2} $$ This of course implies that $$ f(x)=\arctan x+\arccot x $$ is constant. Since $f(0)=\arctan0+\arccot0=\pi/2$, we can conclude that $$ \arctan x+\arccot x=\frac{\pi}{2} $$

Answer 2

$$\frac d{dx}\arctan(x)+\frac d{dx}\mathrm{arccot}(x)=\frac d{dx}\left(\pm\frac{\pi}{2}\right).$$

Now up to you to see when there is a switch between $+$ and $-$.

Answer 3

Since the arccot function (at least in some conventions, like in Mathematica) is only defined for $x \neq 0$, the function $f(x)=\arctan x + \operatorname{arccot} x$ (defined for $x \neq 0$) has the derivative $f'(x)=0$ (defined for $x \neq 0$). A function which has derivative zero is locally constant, so $f(x)=C_1$ for $x<0$ and $f(x)=C_2$ for $x>0$, but $C_1$ doesn't have to be equal to $C_2$.

However, if you use the convention from Wikipedia, then $f(x)$ (defined on the whole real line) is indeed constant on $\mathbf{R}$.